Essay writing: the thesis statement

When you tutor high school English, essay writing is a key topic.  Therefore, the thesis statement is often a focus during English tutoring.

Let’s imagine you write an essay about why camping is great.  (I don’t happen to like camping, but you don’t have to believe in a topic to write a successful essay about it.) You begin with an introductory paragraph that contains a few reflections about camping. From most people’s point of view, the last sentence of the introductory paragraph should be the thesis statement.

Let’s imagine now you’ve finished the essay, your excitement about camping having propelled you through the body paragraphs and conclusion.  You look back.  Here’s what you see:

body paragraph 1:

One great benefit of camping is meeting new people at the campsite.  Since often you’re camped in neighboring spaces…..

body paragraph 2:

Another great feature of camping is being out in the fresh air.  Anyone from the city will quickly notice…..

body paragraph 3:

Yet another great dimension of camping is the family time uninterrupted by work, scheduled activities, or TV shows.  Families often find camping brings them closer….

Now, you look at your introductory paragraph:

introductory paragraph:

Commonly, people love camping.  It is a staple of summer vacations.  Taking afternoon walks along forest paths, roasting hot dogs, and looking up at the starry sky from around the fire are cherished memories of any camper.  Not only must such memories be passed down, but in fact everyone should have them, even if they don’t grow up with a camping tradition.  Usually, people who are skeptical about camping turn out to love it.

With that introductory paragraph, followed by the body paragraphs referred to earlier, the essay will be a “fail”.  Why?  Simple:  it doesn’t have a thesis statement.  An essay that lacks a thesis statement, or else fails to prove its thesis, will get a failing grade.

Since the essay lacks a thesis statement, let’s give it one.  We’ll add a thesis statement at the end of the introduction, as follows:

introductory paragraph with thesis statement added:

Commonly, people love camping.  It is a staple of summer vacations.  Taking afternoon walks along forest paths, roasting hot dogs, and looking up at the starry sky from around the fire are cherished memories of any camper.  Not only must such memories be passed down, but in fact everyone should have them, even if they don’t grow up with a camping tradition.  Usually,people who are skeptical about camping turn out to love it.  There are three great benefits of camping:  meeting new people, being out in the fresh air, and spending time together as a family.

Now, having a thesis statement that mentions the topics in its three body paragraphs, the essay has jumped from a “fail” to a “B” (or thereabouts).   Of course, it needs a conclusion, which will recap the intro and restate the thesis.  Usually, the conclusion is shorter than the introduction.  With a little polish, moving up from “B” to “A” is not too difficult.

The way to fail an essay is to lack a thesis statement or else fail to prove it.  The way to write a B-range essay is to include, in the introductory paragraph, an obvious thesis statement that connects with the body paragraphs that follow.  The way to go from “B” to “A” is with polish.

I’ll have much more to say about essay writing in upcoming posts.  The weekend promises to be sunny and fine; some people are talking about going camping.  Not me:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Arc length: the proportion method

Tutoring math 12, you cover arc length.  The math tutor describes a method students from a generation ago might appreciate.

Arc length is distance along the circumference of a circle.  Following a 360° angle, you of course travel the entire circumference, which is 2πr, r being the radius.

What is you’re traveling less the 360°?  How can you calculate the corresponding arc length?

Example:  Calculate the arc length of a 110° angle on a circle of diameter 15 cm.
Solution: First, realize that half the diameter is the radius. Therefore, the radius of this circle is 7.5cm.

Next, set up the following proportion:

(1)   \begin{equation*}\frac{arc\ length}{angle}=\frac{arc\ length}{angle}\end{equation*}

We know that for 360°, the arc length is 2\pi(r)=2\pi(7.5). Therefore, we adapt the proportion as follows:

(2)   \begin{equation*}\frac{2\pi(7.5)}{360}=\frac{l}{110}\end{equation*}

Next, we invoke the old “cross multiplication” trick described here. It yields

(3)   \begin{equation*}360l=2\pi(7.5)110\end{equation*}

Dividing both sides by 360, we get

(4)   \begin{equation*}l=\frac{2\pi(7.5)110}{360}=14.4\end{equation*}

Apparently the arc length, if we only traverse 110°, is 14.4cm. Given that the arc length would be 2\pi(7.5)=47.1cm for the entire circle, our answer makes sense. 110 is just under a third of 360; correspondingly, 14.4cm is just under a third of 47.1cm.

I’ll be covering another method for arc length soon. Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: the pigeonhole principle

Tutoring math, you might see this topic from a university student.  The math tutor offers it as a point of interest.

The pigeonhole principle is used to solve counting problems.  A simple example:

In a ten day stretch, how many days must it rain to guarantee two consecutive days of rain?

The answer is six, and here’s why:

Step 1:  Organize the numbers from 1 to 10 into sets of two consecutive numbers:

{1,2}, {3,4}, {5,6}, {7,8}, {9,10}

Step 2: Imagine each number represents a day.  If we pick only five numbers, we can pick one from each set, potentially avoiding a consecutive pair (we can pick 1,3,5,7, and 9, for instance).  However, when we pick the sixth one, we must return to one of the five pairs from which we’ve already drawn.  Therefore, we must pick the second number of a consecutive pair, guaranteeing two consecutive days of rain.

Looking at the example above from the point of view of the pigeonhole principle, the pigeons are the numbers we pick, while the holes are the sets of consecutive pairs.  If we have six pigeons, but only five holes whence they came, two must come from the same hole.

The pigeonhole principle can be used to solve some surprising problems.  We’ll look at other examples in upcoming posts:)

Source: Grimaldi, Ralph. Discrete and Combinatorial Mathematics.
  Addison-Wesley,1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: the fourth root of 80

Questions about roots – especially higher order ones – are commonly asked during math tutoring.  The tutor offers a quick example.

 

Example: simplify \sqrt[4]{80}

Step 1: Realize that 80=16(5). Therefore,

(1)   \begin{equation*}\sqrt[4]{80}=\sqrt[4]{16}\sqrt[4]{5}\end{equation*}

Step 2: Realize that the reason we break 80 into 16(5) is that \sqrt[4]{16}=2 (for the reason that 2x2x2x2=16). Therefore,

(2)   \begin{equation*}\sqrt[4]{80}=2\sqrt[4]{5}\end{equation*}

Most fourth roots cannot be simplified in this way; only those that contain a perfect fourth power (one of 2^4=16,\ 3^4=81,\ 4^4=256, etc).

Example: Simplify \sqrt[4]{1250}

Solution:

(3)   \begin{equation*}\sqrt[4]{1250}=\sqrt[4]{625}\sqrt[4]{2}\end{equation*}

\sqrt[4]{625}=5. Therefore,

(4)   \begin{equation*}\sqrt[4]{1250}=5\sqrt[4]{2}\end{equation*}

Thanks for stopping by on this warm summer night:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

English: defense vs defence

Tutoring English, you can’t help but notice nuances.  Here’s one the English tutor has thought about since elementary school.

About 35 years ago I asked my elementary teacher whether to use defence or defense.  She was sporting; we looked in the dictionary together.  It was a British-focused one, so it preferred defence, still allowing defense.  Defense, I think it remarked, was American.

Today I have three different dictionaries in front of me.  Merriam-Webster says defence is a British variation of defense.  Oxford defines defense as “= defence.”  My Collins Essential Canadian English doesn’t have an entry for defense; it only contains defence.

Apparently, defense is American, while defence is British.  Here’s where the American one makes more sense, though:  as far as I can tell, defensive is spelled with an “s”, never with a “c”, in each of the dictionaries.  Possibly, the Yanks went with defense because of the spelling of defensive.

Thanks for coming by.  Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

English: what is a verbal?

Tutoring English, you need to be on your toes.  The tutor is glad he found this term before he was asked about it:)

English is different from the other school subjects in one key way:  the student probably learns more about it outside school than at school.  Very likely, a person might use a certain construction without knowing the theory behind it.

Today I discovered the term verbal.  A verbal is a verb that is being used as a noun, adjective, or adverb.  Some examples:

I love his cooking.   Noun: his cooking is being referred to as the object of someone’s appreciation.

Looking worn, he met us outside the exam room.   Adjective: the verb to look, used in looking worn, is being used to describe a person.

He tried to draw the face.   Adverb: the verb to draw is being used to explain the action of “tried”.

I confess that, waking up this morning, I had never heard the word verbal. Now we know:)

Thanks for stopping by on this beautiful day. Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Source: Harbrace Handbook for Canadians, 6th edition. 2003: Nelson Education Ltd.

Math: a mean question

Continuing from last post, the tutor gives an example of a question involving the mean.

You will often find a question such as this on a qualifying exam:

Bill writes six math tests. They all have equal weighting towards his final mark. Suppose, after five tests, his average (mean) mark is 71, yet he needs a final average of at least 73. What mark must he manage on his sixth test?

Solution:

Let x be the mark he needs on his sixth test. If his average after five tests is 71, the equivalent situation is that he got 71 on each of those five. Our general formula for average (mean) is as follows:

(1)   \begin{equation*}mean=\frac{sum\ of\ all\ values}{number\ of\ values}\end{equation*}

In Bill’s case,

(2)   \begin{equation*}mean=\frac{71+71+71+71+71+x}{6}=73\end{equation*}

which simplifies to

(3)   \begin{equation*}\frac{355+x}{6}=73\end{equation*}

Multiplying both sides by 6 gives

(4)   \begin{equation*}355+x=438\end{equation*}

Subtract 355 from both sides, giving

(5)   \begin{equation*}x=83\end{equation*}

Apparently, Bill needs 83 on his sixth test to bring his average from 71 to 73. Here’s one way to see it: he needs 73 on his last, plus 2 points for each of the five he got 71 on. 73+5(2)=83.

I’ll be further discussing the mean with applications in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Mean, median, and mode

Tutoring high school math, you deal with introductory statistics.  The tutor introduces measures of central tendency.

Let’s imagine you have the following seed counts from eleven different apples:

15 9 8 10 8 10 8 5 7 14 12

We wish to find the mean, median, and mode of the seed counts.

When I was a kid, people said “average” instead of mean. To find the mean, we add all the numbers together, then divide by how many there are (in this case, 11):

(1)   \begin{equation*}mean=\frac{(15+9+8+10+.....+14+12)}{11}=9.6363...\end{equation*}

So, the mean number of seeds in the apples is 9.6363....

To find the median, you line up the numbers from least to greatest. The median is the middle one.

We rewrite the numbers in ascending order:

5 7 8 8 8 9 10 10 12 14 15

The middle number is 9. Therefore, the median number of seeds in the apples is 9.

The mode is simply the count that occurs most often. In this case, it’s 8. At three times, 8 happens more than any other count.

Some points to note:

  1. At 9.6363....., the mean doesn’t actually occur in the data set. Very often, such is the case.
  2. What if the list has an even number of entries? An odd-numbered list has a clear “middle” one, but an even-numbered list doesn’t.
  3.  
    Consider the following list of six entries:

    5 6 7 10 13 14

    Here, the median is \frac{7+10}{2}=8.5. To find the median of an even list, you take the two middle values, add them, then divide by two.

  4. What if, as in the six-membered list just above, there is no “most frequent” value? What is the mode in such a case?
  5.  
    For the six-membered list just above, you can see it in two ways: either there’s no mode, or else there are six different modes. However, consider the following list:

    1 1 2 3 5 6 7 7 9

    For this list, we would say there are two modes: 1 and 7.

The mean, median, and mode are all called measures of central tendency: they all estimate what the “next” value would be. I’ll be saying more about them in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: simplification with a rational exponent

Tutoring math, this topic is perennial.  The math tutor works an example.

Imagine you are posed the following question:

    \[ Simplify\ 32^{\frac{6}{5}}\]

First, realize the following:

    \[32^{\frac{6}{5}}=\sqrt[5]{32}^6\]

\sqrt[5]{32}=2, since 2*2*2*2*2 = 2^5 = 32. Therefore,

    \[32^{\frac{6}{5}}=\sqrt[5]{32}^6=2^6\]

2^6 = 2*2*2*2*2*2 = 64. So, from start to finish, we have

    \[32^{\frac{6}{5}}=\sqrt[5]{32}^6=2^6=64\]

I’ll be saying more about this topic in future posts. Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.